Download Lecture 6: End and cofinal extensions

MODEL THEORY OF ARITHMETIC
Lecture 6: End and cofinal extensions
Tin Lok Wong
12 November 2014
A relatively neglected aspect of the study of nonstandard models of arithmetic is the
study of their cofinal extensions. These extensions certainly do not present themselves
to the intuition as readily as do their more popular cousins the end extensions; but
they are not exactly shrouded in mystery or unnatural objects of study either. They
are equal partners with end extensions in the construction of general extensions of
models; they offer both special advantages and disadvantages worthy of our interest;
and, occasionally, they are useful in understanding the generally more simply behaved
end extensions.
Craig Smoryński [16]
Recall BΣn+1 ` IΣn for every n ∈ N from Theorem 2.10. The aim of this lecture is to show
that actually these theories are not very far apart.
Theorem 6.1 (Harvey Friedman, Jeff Paris [12], independently). Let n ∈ N. Then BΣn+1 is
Πn+2 -conservative over IΣn , i.e., for every σ ∈ Πn+2 ,
BΣn+1 ` σ
⇒
IΣn ` σ.
The particular case when σ = ⊥ is essentially the equiconsistency between BΣn+1 and IΣn .
However, this conservation theorem tells us much more than just equiconsistency. For example,
establishing the equiconsistency between BΣn+1 and IΣn in ZFC is trivial, because ZFC proves
the consistencies of both BΣn+1 and IΣn . Contrarily, this conservation result, even established
within ZFC, really has non-trivial content.
Note that IΣn+1 is never Π1 -conservative over the corresponding BΣn+1 by Gödel’s Second
Incompleteness Theorem, because IΣn+1 ` Con(BΣn+1 ). A proof of this can be found in Section I.4
of the Hájek–Pudlák book [8].
The proof of Theorem 6.1 we present here is a simplification of that in Clote–Hájek–Paris [4],
using end and cofinal extensions. Recall
K ⊇e M
means
∀k∈K \ M ∀m∈M m 6 k,
K ⊇cf M
means
∀k∈K \ M ∃m∈M k 6 m.
We split into two threads, which will merge in the end to give the conservation theorem.
6.1
The strength of end extensions
We may define BΠn in a way analogous to the definition of BΣn , but as mentioned in Proposition 3.8, we get no new theory out of this. We prove the proposition in full here.
Proposition 6.2. BΣn+1 and BΠn are equivalent for all n ∈ N.
Proof. If ∀x<a ∃y ∃v ψ(v, x, y, z̄), where ψ ∈ Πn , then applying BΠn gives
∃b ∀x<a ∃y, v<b ψ(v, x, y, z̄),
so that ∃b ∀x<a ∃y<b ∃v ψ(v, x, y, z̄) in particular.
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K
b
M
↓ underspill
M
a
y
ϕ
x
Figure 6.1: How end extensions imply collection
We need also an upside-down version of overspill.
Underspill. Let n ∈ N and I (e M |= IΣn . If θ ∈ Πn (M ) such that
M |= θ(x)
for all x ∈ M \ I,
then M |= θ(x) for arbitrarily large x ∈ I.
Proof. Otherwise x > b ∧ θ(x) defines M \ I for some b ∈ I, contradicting LΠn .
The collection schemes are intimately connected with end extensions. In general, if a model of
arithmetic has a proper end extension, then it satisfies some collection. The amount of collection
it satisfies depends on what theory the extension has, and how elementary the extension is.
Definition. Let M, K be LA -structures and n ∈ N. Then M is a Σn -elementary, or simply
n-elementary, substructure of K, written M 4n K, if M ⊆ K and for all θ ∈ Σn and all c̄ ∈ M ,
M |= θ(c̄)
⇔
K |= θ(c̄).
In this case, we informally say that such θ(c̄)’s transfer between M and K.
Notice requiring θ ∈ Πn instead does not change the notion defined.
Theorem 6.3 (Adamowicz–Clote–Wilkie [3]). Fix n ∈ N. Let M |= PA− and K n,e M satisfying IΣn . Then M |= BΣn+1 .
Proof. Notice M |= I∆0 by Corollary 5.1. In view of Proposition 6.2, it suffices to show M |= BΠn .
Let a ∈ M and ϕ ∈ Πn (M ) such that M |= ∀x<a ∃y ϕ(x, y). Consider
{b ∈ K : K |= ∀x<a ∃y<b ϕ(x, y)}.
| {z }
Πn
|
{z
Πn over K |= BΣn
}
It includes K \ M because M 4n,e K. So by Πn -underspill, we find b ∈ M such that
K |= ∀x<a ∃y<b ϕ(x, y).
This transfers down to M since M 4n,e K.
The case n = 0 of this theorem strengthens Corollary 5.1 for proper cuts.
6.2
The Splitting Theorem
We need a hierarchical version of the Tarski–Vaught Test [17]. The significance of this test is that
it allows us to talk about elementarity without reference to what is true in the smaller model. This
is especially helpful when we do not know much about the smaller model, for example, when it is
still being constructed.
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Tarski–Vaught Test. Let M, K be LA -structures, and suppose M 40 K. Then the following
are equivalent for all n ∈ N.
(a) M 4n+1 K.
(b) For every η(x̄) ∈ Πn (M ), if K |= ∃x̄ η(x̄), then K |= η(c̄) for some c̄ ∈ M .
Proof. The implication (a) ⇒ (b) is straightforward. We prove the converse implication (b) ⇒ (a)
by strong induction on n.
Let n ∈ N for which (b) is true, and (b) ⇒ (a) for all smaller indices. Then we know M 4n K,
either from the assumption M 40 K when n = 0, or from the induction hypothesis when n > 0.
Consider the Σn+1 (M )-formula ∃x̄ η(x̄), where η ∈ Πn (M ). Clearly M |= ∃x̄ η(x̄) implies K |=
∃x̄ η(x̄) since M 4n K. If K |= ∃x̄ η(x̄), then
K |= η(c̄)
for some c̄ ∈ M
by (b)
∴
M |= η(c̄)
for some c̄ ∈ M
since M 4n K
∴
M |= ∃x̄ η(x̄).
Another fact that we need is an alternative axiomatization of the induction schemes in terms
of strong collection. Recall that collection says all functions with bounded domains have bounded
images. Strong collection says that all partial functions with bounded domains, not only total
ones, have bounded images. This gives yet another way of showing BΣn+1 ` IΣn for every n ∈ N.
Theorem 6.4 (Harvey Friedman). The following are equivalent over I∆0 for all n ∈ N.
(a) IΣn+1 .
(b) Strong Πn -collection: for every θ ∈ Πn ,
∀z̄ ∀a ∃b ∀x<a ∃y η(x, y, z̄) → ∃y<b η(x, y, z̄) .
Proof. For (a) ⇒ (b), use Σn+1 -separation from Theorem 2.7 to find c such that
Ack(c) = {x < a : ∃y η(x, y, z̄)}.
Recall IΣn+1 ` BΣn+1 from Theorem 2.2. So
∃b ∀x<a ∃y<b x ∈ Ack(c) → η(x, y, z̄) .
| {z }
| {z }
∆0
|
Πn
{z
}
Πn
We show (b) ⇒ (a) by strong induction on n. Let n ∈ N such that strong Πn -collection holds,
and (b) ⇒ (a) for all smaller indices. Then we get IΣn , either from the base theory I∆0 when
n = 0, or from the induction hypothesis when n > 0. As in the proof of Theorem 5.9, it suffices
to show that every nonempty bounded Σn+1 -definable set has a minimum. Fix a and consider the
Σn+1 -formula ∃y η(x, y, z̄), where η ∈ Πn . Strong Πn -collection gives b such that
{x < a : ∃y η(x, y, z̄)} = {x < a : ∃y<b η(x, y, z̄)}.
| {z }
Πn
|
{z
Πn over BΣn
}
This set is Πn -definable because either n = 0 and BΣn is not needed, or n > 0 and we have BΣn
from IΣn . So provided it is nonempty, it must have a minimum by LΠn .
Remark 6.5. Similar to Proposition 6.2, strong Σn+1 -collection, appropriately defined, is the same
as strong Πn -collection for every n ∈ N.
We are now ready for the Splitting Theorem, which says that every extension splits into a cofinal
extension followed by an end extension. Clearly, to split an extension in such a way, the middle
model must consist of the downward closure of the ground model.
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K <n+1 M
K<M
× new
M |= IΣn
M |= IΣn
M = supK M
⇒ M 4n+1 M 4n K
M |= BΣn+1
Figure 6.2: The Splitting Theorem
Figure 6.3: Constructing cofinal extensions
using the Splitting Theorem
Definition. If S ⊆ M |= PA− , then
supM S = {x ∈ M : x < s for some s ∈ S}.
More importantly, some elementarity is preserved in such splits. This should be surprising
because the supremum of a model is an entirely order-theoretic notion, and it is not clear at first
glance why it should have any bearing on the arithmetic structure.
Splitting Theorem (essentially Kaye [9]). Fix n ∈ N. Let M |= IΣn and K <n+1 M . Then
M 4n+1,cf M 4n,e K, where M = supK M .
Proof. We follow the proof from Enayat–Mohsenipour [5].
We first show the n-elementarity between M and K. Note M ⊆e K. So M 40 K by Proposition 4.12. Thus we are already done if n = 0. Now, suppose n = m + 1. Then the Tarski–Vaught
Test applies. Let η(x, y) ∈ Πm and c ∈ M such that K |= ∃y η(c, y). Take a ∈ M strictly above c,
which is possible since M ⊆cf M . Using strong Πm -collection, find b ∈ M such that
M |= ∀x<a ∃y η(x, y) → ∃y<b η(x, y) .
| {z }
| {z }
Πm
|
{z
Σm+1
|
Πm
}
{z
Πm+2
|
{z
Σm+1
}
}
This transfers to K by (m + 2)-elementarity, making K |= ∃y<b η(c, y). Any witness to this must
be in M because b ∈ M .
Let us move on to the (n + 1)-elementarity between M and M . First, we know M 4n M
since M 4n+1 K <n M . Let η(x) ∈ Πn (M ). Then M |= ∃x η(x) implies M |= ∃x η(x) because
M 4n M . Conversely, if M |= ∃x η(x), then
∴
K |= ∃x η(x)
since M 4n K
M |= ∃x η(x)
since M 4n+1 K.
The Splitting Theorem tells us that end extensions and cofinal extensions are essentially the
only interesting kinds of extensions for models of arithmetic, because every other extension factors
into these.
There are a number of other splitting theorems in the model theory of arithmetic in the literature. We will meet one more in the Further exercises.
6.3
Cofinal extensions
We are only one step away from the conservation result. The stepping stone is, nevertheless, of
independent model-theoretic interest. Its countable version was first proved in Paris [12].
Theorem 6.6. Let n ∈ N and M |= IΣn . Then there is M <n+1,cf M that satisfies BΣn+1 .
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Proof. Use Compactness to find K < M such that M 6⊆cf K. Let M = supK M . The Splitting
Theorem implies M n,e K and M 4n+1,cf M . So M |= BΣn+1 by Theorem 6.3.
This proof actually shows the stronger fact that the supremum of a model of IΣn in every
non-cofinal extension is a model of BΣn+1 .
Theorem 6.6 readily implies the Friedman–Paris conservation theorem, because if BΣn+1 proves
much more than what IΣn does, then there must be a model of IΣn which deviates from BΣn+1
so much that it cannot fit inside any model of BΣn+1 .
Proof of Theorem 6.1. Let σ ∈ Πn+2 such that IΣn 0 σ. Pick M |= IΣn + ¬σ. Find M <n+1 M
satisfying BΣn+1 using Theorem 6.6. Then M |= ¬σ because ¬σ ∈ Σn+2 . So BΣn+1 0 σ.
Further exercises
The MRDP Theorem, mentioned in the Further reading section of Lecture 1, says that every
Σ1 -formula is uniformly equivalent to an existential LA -formula in N. A careful check reveals that
its proof can actually be formalized within I∆0 + exp. So we informally write I∆0 + exp ` MRDP.
Fact 6.7 (Gaifman–Dimitracopoulos [7]). For every Σ1 -formula
θ(x̄), there exists an existential
LA -formula θ0 (x̄) such that I∆0 + exp ` ∀x̄ θ(x̄) ↔ θ0 (x̄) .
The aim of these exercises is to prove the Gaifman Splitting Theorem [6, 7] assuming MRDP
holds in I∆0 + exp. The main difference between our Splitting Theorem and Gaifman’s is that his
theorem does not start with any elementarity.
Gaifman Splitting Theorem. If M, K |= PA such that M ⊆ K, then M = supK M < M .
(a) Recall ∆0 ⊆ Σ1 ∩ Π1 . Show that if M, K |= I∆0 + exp and M ⊆ K, then M 40 K.
(b) Let θ be an LA -formula and M ⊆cf M |= PA− . Suppose for every a ∈ M , there exists b ∈ M
such that M |= ∀x<a ∃y<b θ(x, y). Explain why M |= ∀x ∃y θ(x, y).
(c) Using (b), or otherwise, show that if M |= PA− + Coll(Σ1 ) and M <0,cf M satisfying PA− ,
then M 42 M .
(d) Fix n ∈ N. Follow the steps below to show that if M |= BΣn+1 + exp and M <n+1,cf M ,
then M <n+2 M .
Let θ ∈ Πn (M ) such that M |= ∀x ∃y θ(x, y). Since M ⊆cf K, it suffices to show K |=
∀x<a ∃y θ(x, y) for every a ∈ M . Pick a ∈ M . Define f : {x ∈ M : x < a} → M by setting
f (x) = min{y ∈ M : M |= θ(x, y)}.
(i) Explain why f is well-defined and is coded in M .
(ii) Show that the code of f in M also codes a total function {x ∈ K : x < a} → K in K.
(iii) Explain why K |= ∀x<a θ(x, f (x)).
(e) Combine (c) and (d) to show that if M |= PA and M <0,cf M satisfying PA− , then M < M .
(f) Deduce the Gaifman Splitting Theorem from (a) and (e). Observe that the bigger model
actually does not need to satisfy full PA: having I∆0 + exp is sufficient.
Further comments
More about the Friedman–Paris conservation theorem
Many proofs of Theorem 6.1 are known. The first one by Paris [12] involves his hierarchy of cuts.
Later he gave a simpler proof using definable ultrapowers [13]. Other model-theoretic proofs use
Skolem hulls [10, Chapter 10] or Herbrand saturated models [1]. Several proof-theoretic proofs are
known too [2, 15].
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The Πn+2 -conservativity of BΣn+1 over IΣn is actually provable in I∆0 + supexp, where supexp
is an axiom asserting the existence of
. . .2
x-many 2’s
22
for every x. See the paper by Clote, Hájek, and Paris [4] for more information.
Collection and end extensions
The theory BΣn+1 we obtained in Theorem 6.3 is best possible. A lot is known about similar
equivalences between end extendability and strength. These results can be summarized as follows.
T
Definition. Let k, n ∈ N. Then ThIΣn (k-ecuts) = {Th(I) : I k,e M |= IΣn }. We denote by
Πk -Cn(IΣn ) the Πk -consequences of IΣn .
Theorem 6.8. Let n, k ∈ N. Then
(
BΣk+1 + Πk+1 -Cn(IΣn ) if k 6 n + 1;
ThIΣn (k-ecuts) =
BΣk
if k > n + 2.
Moreover, every model of ThIΣn (k-ecuts) is elementarily equivalent to some I k,e M |= IΣn .
The case when k > n + 1 can be extracted from Paris–Kirby [14]. One direction of the case
when k 6 n is provided by our Theorem 6.3. The other direction can be proved by following
Solovay’s self-embedding argument [13] when the ground model itself satisfies IΣn , or by following
the omitting types argument in Paris–Kirby [14] otherwise.
Notice Theorem 6.8 does not say that every model of ThIΣn (k-ecuts) is a proper k-elementary
cut of some model of IΣn . Apparently, some extra saturation is needed to guarantee the existence
of an end extension of the appropriate kind, but currently, the exact amount of saturation required
is mostly unknown. Here we gather the key questions.
Question 6.9 (Wilkie–Paris [19]). Does every countable model of BΣ1 have a proper end extension
satisfying I∆0 ?
Question 6.10 (Clote [3]). Let n ∈ N. Does every model of BΣn+2 have a proper (n + 2)elementary end extension satisfying I∆0 ?
Question 6.11 (Clote [3], after Matt Kaufmann). Let n ∈ N. Does every countable model
of BΣn+2 have a proper (n + 2)-elementary end extension satisfying BΣn+1 ?
Wilkie and Paris [19] showed that Questions 6.9 and 3.9 cannot both have a positive answer.
Formalizing the MRDP Theorem
There is a sense in which MRDP Theorem is equivalent to the Gaifman Splitting Theorem, see
Exercise 7.6 in Kaye’s book [3]. It is open whether Fact 6.7 remains true without exp.
Question 6.12 (Jeff Paris). Does I∆0 ` MRDP?
Wilkie [18] noticed that a positive answer to this question implies NP = co-NP.
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